A305048 Number of ordered pairs (k, m) of nonnegative integers such that 5^k + 10^m is not only a primitive root modulo prime(n) but also smaller than prime(n).

0, 1, 1, 0, 2, 3, 2, 2, 2, 4, 1, 3, 5, 1, 4, 3, 3, 4, 2, 2, 3, 2, 4, 4, 2, 5, 4, 4, 2, 2, 2, 4, 4, 6, 6, 4, 3, 3, 7, 6, 6, 2, 4, 3, 5, 3, 2, 3, 8, 3, 4, 4, 1, 3, 5, 5, 6, 5, 6, 4, 3, 5, 1, 1, 3, 4, 4, 2, 7, 2, 4, 4, 2, 8, 3, 7, 7, 3, 5, 4, 6, 1, 3, 4, 4, 7, 5, 4, 6, 2

Offset: 1

Author

Zhi-Wei Sun, May 24 2018

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 4. In other words, any prime p > 7 has a primitive root g < p of the form 5^k + 10^m with k and m nonnegative integers.

We have verified this for any prime p > 7 not exceeding 10^9.

It seems that a(n) = 1 only for n = 2, 3, 11, 14, 53, 63, 64, 82, 99, 101, 111, 129, 344, 369, 391, 795, 1170, 1587, 5629, 5718, 6613, 430516.

See also A305048 for similar conjectures.

Examples

			a(14) = 1 with 5^2 + 10^0 = 26 a primitive root modulo prime(14) = 43.
a(101) = 1 with 5^0 + 10^0 = 2 a primitive root modulo prime(101) = 547.
a(111) = 1 with 5^2 + 10 = 35 a primitive root modulo prime(111) = 607.
a(5718) = 1 with 5^0 + 10^3 = 1001 a primitive root modulo prime(5718) = 56401.
a(6613) = 1 with 5^1 + 10^3 = 1005 a primitive root modulo prime(6613) = 66301.
a(430516) = 1 with 5^5 + 10^1 = 3135 a primitive root modulo prime(430516) = 6276271.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..100000
• Zhi-Wei Sun, New observations on primitive roots modulo primes, arXiv:1405.0290 [math.NT], 2014.
• Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28--Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

A000040, A000351, A011557, A303540, A239957, A241476, A241504, A241516, A305030.

Programs

Mathematica

p[n_]:=p[n]=Prime[n];
Dv[n_]:=Dv[n]=Divisors[n];
gp[g_,p_]:=gp[g,p]=Mod[g,p]>0&&Sum[Boole[PowerMod[g,Dv[p-1][[k]],p]==1],{k,1,Length[Dv[p-1]]-1}]==0;
tab={};Do[r=0;Do[If[gp[5^a+10^b,p[n]],r=r+1],{a,0,Log[5,p[n]-1]},{b,0,Log[10,p[n]-5^a]}];tab=Append[tab,r],{n,1,90}];Print[tab]